Environmental and Exploration Geophysics I

tom.h.wilsonwilson@geo.wvu.edu

Department of Geology and GeographyWest Virginia University

Morgantown, WV

Magnetic Magnetic Methods (III)Methods (III)

To simplify this problem assume that the anomaly arises from a vertically polarized sphere. Assume Z = FA and compute FAT for a point directly over the center of the chamber.

Could a total-field magnetic survey detect the illustrated burial chamber (spherical void) in a region where FE = 55,000nT and i = 70o?

2/522

22

)(

)2(

zx

xzMZA

See equation 7-35

2/522

22

)(

)2(

zx

xzMZA

Equation 7-35 provides a nice short cut approach that allows us to avoid some of the complexity in equations 7-36 and 7-37.

Remember what M is?

M

and also recall I and so M VIV

M is the magnetic dipole moment.

Do you recall the relationship between I and FE?

E EI kF so M VkF

34

3 EM R kF

2 2

2 2 5/ 2

(2 )

( )A

M z xZ

x z

Let’s use this simpler form to answer Question 7.5.

We have another dilemma! How good are equations 7-36 and 7-37? Can we sort out the truth?

Berger notes that the source of his equations is Telford et al. (Applied Geophysics). Examination of Telford’s text reveals that Berger’s equations are different. Telford’s equations for Z and H are as follows.

232

3 2 2 5/ 2

4 2 3cos 1 cot3

( )

E

A

x xR kF i i zzHz x z

232

3 2 2 5/ 2

4sin 2 3 cot

3( )

E

A

x xR kF i iz zZz x z

In our study of gravity methods, we developed analytical expressions for the gravitational effects of simple geometrical objects like the sphere, horizontal cylinder and flat plate. We can take the same approach with magnetic methods. Simple geometrical objects can often be used to approximate the effects of more complicated objects such as dikes, sills, faulted layers, mine shafts, cavities, caves, etc.

Estimates of maximum depth, susceptibility contrast, size of objects with a given susceptibility, etc. can often be estimated quickly and without the aid of a computer.

2/522

22

)(

)2(

zx

xzMZA

The Vertical Magnetic Anomaly over a Vertically Polarized Sphere

2/522

223

)(

)2(34

zx

xzIRZA

2/522

223

)(

)2(34

zx

xzkFRZ

E

A

See Handout

2/52

25

2

223

)1(

)2(34

z

xz

z

xzkFR

ZE

A

Pull the z2 out of the terms in parenthesis to yield

2/52

2

2

2

3

3

)1(

)2(34

z

xz

x

z

kFRZ

E

A

Note that when x=03

3

max38

z

kFRZ

E

3

3

max38

z

kFRZ

E

is the maximum value of the anomaly.

2/5

2

2

2

2

max1

2

2

1)(

z

x

z

x

Z

xZA

The ratio Z/Zmax describes the relative variations of the magnetic anomaly produced by a vertically polarized sphere as a function of the distance variable x/z, where z is the depth to the center of the polarized sphere. The equation is independent of the object’s size and susceptibility and the magnitude of the earth’s main field FE.

Vertical Field - i=90, R=0.4, z=1.75, F=43000

-1

0

1

2

3

4

5Z

V (

nT

)

-6 -4 -2 0 2 4 6

x (meters)

For Z/Zmax=0.5

X1/2~0.875

implies Z=?

For Z/Zmax=0

X0~ 2.25

implies Z=?

Using a z of 1.75 meters ..

Z=2X1/2

Vertically Polarized Sphere

The “x” variable represents surface distance from the anomaly peak in terms of the depth to the center of the sphere (z). Just as we did in our analysis of the variations of g over simple geometrical objects, we let the ratio Z/Zmax equal to the value at which the anomaly falls from its peak to 1/2 its maximum value (i.e. Z/Zmax = 1/2), for example, and determine at what distance x/z, the anomaly falls to 1/2 of its maximum value.

The table below lists diagnostic position, the value of x/z at which the anomaly drops to a given diagnostic value and the depth-index multiplier used to convert x to z (I.e. just (x/z)-1).

Vertical Field - i=90, R=0.4, z=1.75, F=43000

-1

0

1

2

3

4

5Z

V (

nT

)

-6 -4 -2 0 2 4 6

x (meters)

For Z/Zmax=0.5

X1/2~0.875

implies Z=?

For Z/Zmax=0

X0~ 2.25

implies Z=?

Using a z of 1.75 meters ..

-

Vertically Polarized Vertical Cylinder Vertical Component ZA

A vertically oriented dipole with one pole located at a great distance has the magnetic field of an isolated monopole.

Refer to equation 7-22 in Burger. As Burger illustrates, ZA is derived by evaluating

dz

dVZA

If we assume r lies in the xz plane then this derivative becomes

2/322 )( zx

AzkFZ E

A

recall m

Vr

2/322

2

2/322 )(or ,

)( zx

zIR

zx

AzkFZ E

A

as shown on your handout. Recall I=kFE.

We can simplify the above equation to

2/32

23

2

)1(

z

xz

zIRZA

2/32

22

2

)1(

1

z

xz

IRZA

3

223/ 2

2

4 13

( 1)sph

G Rg

xzz

2/32

22

2

)1(

1

z

xz

IRZA

From this equation -

we see that - 2

2

maxz

IRZ

2/32

2max )1(

1

z

xZ

ZA

and also that -

23/ 2max

2

1

( 1)

sphg

xgz

Vertically Polarized Vertical Cylinder

2/32

2max )1(

1

z

xZ

ZAAgain, from this normalized or relative form of the equation for the vertical anomaly over the isolated pole that we can solve for relative value of ZA/Zmax in

terms of the variable x/z. Thus when we examine an anomaly in real data, we can relate the position, X1/2 (the distance from the peak anomaly to the point where it falls to 1/2 its maximum value), for example, to the depth z, where z Is calculated from X1/2 or other diagnostic positions using the depth-index multipliers.

As we would expect, these values are identical to those obtained for the variations in g observed over a buried sphere (e.g. monopole!).

0

1

2

3

4

5

6

7

8

Z (nT)

Vertical Cylinder (monopole)

-6.0 -4.0 -2.0 0.0 2.0 4.0 6.0

x (meters)

X1/2

X1/2 ~ 1.34 implies Z = 1.31 x 1.34 ~1.75

Determine which anomaly is associated with a buried sphere (dipole) and which arises from a vertical cylinder (or monopole). Also determine their depths (depth to center or depth to top).

Which estimates of Z are more consistent? Compute the range or standard

deviation.Which model - sphere or vertical cylinder - yields the smaller range or

standard deviation?

Anomaly B: Sphere, Vertical Cylinder; Z = __________?

2

2

2

2

2

2

2

1

12

zx

zx

z

IRZ

Vertically Polarized Horizontal Cylinder

2

2

max2

z

IRZ

2

2

2

2

2

max 1

1)(

zx

zx

Z

xZ

Vertically Polarized Horizontal CylinderDiagnostic position

Z/Zmax

x/z (x/z)-1

Depth Index multiplier9/10 0.188 5.323/4 0.31 3.232/3 0.37 2.71/2 0.495 2.021/3 0.61 1.641/4 0.68 1.470 1.0 1

-6 -4 -2 0 2 4 6

x (meters)

Horizontal Cylinder vs. Vertical Cylinder

-4

0

4

8

12

16

Z (

nT

)

Horizontal

Vertical

Sphere vs. Horizontal Cylinder

-4

0

4

8

12

16

Z (

nT)

-6 -4 -2 0 2 4 6

x (meters)

Horizontal Cylinder

sphere

Coming back to question 7-3, we could have used the simplified form of the horizontal cylinder to rapidly estimate the possibility of detecting the buried wall. In our previous evaluation, we might have ruled out the survey if we had made our decision using predictions based on the effect of the vertically polarized sphere.

nTz

IRZ

11.14

22

2

max

Vertically Polarized Faulted Horizontal Slab or Semi Infinite Sheet

22 zx

xItZ

2

22

1z

x

x

z

ItZ

22max )(zz

zItzxZZ

z

ItZ

2max

Horizontal Slab

-8

-6

-4

-2

0

2

4

6

8

Z (nT)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

x (meters)

z t

nTz

ItZ 14.6

2max

Xmax=zz = 1.75m

t = 0.5m

Vertically Polarized Faulted Horizontal Slab or Semi Infinite sheet

2

2max 1

2

z

x

x

zZ

Z

The magnetic response of a sheet of dipoles is obtained by carrying out integrations over two sheets: one consisting of the negative poles and the other of the positive poles.

/ 2

/ 22 cos 2 2

topsheetA

IdyZ Id I

r

where mI area

See Equation 7-45, on page 429 of Berger.

IZtopsheetA 2

The effect of the bottom sheet will also equal

IZsheetbottomA 2

The negative

Sign comes from the convention that defines upward pointing vectors (from the positive pole) as negative. So the net result ….. is

022 nfinite

IIZsheetiA

The process yields an intermediate more useful result.

The contribution from the top of the rod is topI2 and

the contribution from the base of the rod is botI2

The total field of this infinitely long intrusive (dike) will be

bottopA IIZ 22

or just

bottopA IZ 2

Determine the vertical field anomaly (ZA) over the intrusive shown in the diagram (see text) at a point directly over the center of the intrusive. The intrusive has a very long strike-length. FE is vertical and equal to 55,000nT. Use equation 7-46 and compute ZA for two cases. Case 1: assume that the base of the intrusive is located at 12km beneath the surface. Case 2: assume the base is located at infinity. Compare the two results.

bottopA IZ 2

bottopA IZ 2

2A topZ I

Determining

Comparative estimates of Z are used to determine the geometry of the buried object. However, if you are pretty sure about the geometry of the buried object, you may use only one of the diagnostic positions, for example, the half-max (X1/2) distance associated with an anomaly to quickly estimate object depth. If the object were a sphere or buried vertical cylinder , you would use the appropriate half-max relationship.

Burger limits his discussion to half-maximum relationships.

Breiner, 1973

Determine the depth z to the center of the basalt flow. Also indicate whether you think the flow is faulted (two offset semi-infinite sheets) or just terminates (a semi-infinite sheet). What evidence do you have to support your answer? Refer to illustration on page 433 and associated discussion.

tantan2 11

tz

x

z

xkFZ EA

This problem relies primarily on a qualitative understanding of equation 7-47.

Field of the semi-infinite plate

X = 0 at the surface point directly over the edge of the plate. The field at a point X is derived from the two angles shown below - 1 and 2 - used in the text.

1

2

z

t

The angle subtended by the top of the sheet at x is top

2

The angle subtended by the bottom of the sheet at x is bot

2

212)( IxZ sheethalfA

11 tan

2top

x

z

12 tan

2bot

x

z t

tantan2 11

tz

x

z

xIZA

47-7 Eqn. tantan2 11

tz

x

z

xkFZ EA

-4

-3

-2

-1

0

1

2

3

4

Z (nT)

Variations of Z across edge of isolated sheet.

-30 -20 -10 0 10 20 30

x (meters)

Half-plate (the Slab, semi infinite plate, the half-sheet …)

- - - - - - - - - - - - - - - - - - - - - - -+ + + + + + + + + + + + + + + +

z

t

tantan2 11

tz

x

z

xkFZ EA

combinedresponse

uppersheet

lowersheet

-4

-3

-2

-1

0

1

2

Z (nT)

3

4

Semi-Infinite Sheets

-30 -20 -10 0 10 20 30

x (meters)

Look carefully at the anomaly profile shown in Problem 7-8 and consider the overall shape of the anomaly and how it may allow you to discriminate between the faulted versus terminated flow interpretations.

The final will be comprehensive, but the focus will be on material covered since the last exam.